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\(\frac{x}{10}\)^2019+(0,1)^2019=\(\frac{1}{10}\)^2019 \(\frac{x}{10}\)^2019 =(0,1)^2019-(0,1)^2019 \(\frac{x}{10}\)^2019 =0 \(\frac{x}{10}\)^2019 =0^2019 \(\frac{x}{10}\) =0 x =0
#)Giải :
Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\frac{a^{2019}+b^{2019}}{c^{2019}+d^{2019}}\left(1\right)\)
Lại có : \(\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\left(\frac{a}{c}\right)^{2019}=\left(\frac{b}{d}\right)^{2019}=\left(\frac{a+b}{c+d}\right)^{2019}=\frac{\left(a+b\right)^{2019}}{\left(c+d\right)^{2019}}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{\left(a+b\right)^{2019}}{\left(c+d\right)^{2019}}=\frac{a^{2019}+b^{2019}}{c^{2019}+d^{2019}}\left(đpcm\right)\)
\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)
\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x>y\)
x2019-2019.x2018+2019.x2018+2019.x2017-2019.x2016+......2019.x-200 Tại x=2018
Giúp mik vs nhé
Sai đề nên t sửa luôn nhé!
Vì \(x=2018\Rightarrow2019=2018+1=x+1\)
\(A=x^{2017}-2019\cdot x^{2018}+2019\cdot x^{2017}-2019\cdot x^{2016}+....+2019\cdot x-200\)
\(\Rightarrow A=x^{2019}-\left(x+1\right)x^{2018}+\left(x+1\right)x^{2017}-\left(x+1\right)x^{2016}+....-\left(x+1\right)x^2+\left(x+1\right)x-200\)
\(\Rightarrow A=x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-x^{2017}-x^{2016}+....-x^3-x^2+x^2+x-200\)
\(\Rightarrow A=x-200=2018-200=1818\)